Animated Solution for Physics - Thermodynamics: The net work on the gas in the cycle ABCDA is (see above figure)

Welcome to a classic thermodynamics problem that has tricked thousands of students! We are presented with a cyclic process $ABCDA$, but there is a major catch: the graph provided is a pressure-temperature ($p-T$) diagram, not the usual pressure-volume ($p-V$) diagram.

Many students immediately try to calculate the area inside the rectangle, thinking it will give them the net work done. Do not fall for this trap! The area under a cycle only represents work on a $p-V$ diagram.

To find the true net work done, we must roll up our sleeves and calculate the work done in each of the four individual processes: $W_{\text{net}}=W_{AB}+W_{BC}+W_{CD}+W_{DA}$. Let's break them down one by one.

Let's start with the horizontal lines on our graph, which represent processes occurring at a constant pressure. These are isobaric processes.

For process $A\to B$, the pressure is constant at $2\times 10^5\text{ Pa}$, and the temperature increases from $300\text{ K}$ to $500\text{ K}$. The work done in an isobaric process is given by the elegant formula: $W=nR\Delta T$

Substituting our values ($n=2$), we get: $W_{AB}=2R(500-300)=400R$

Now, let's look at the other isobaric process, $C\to D$. Here, the pressure is constant at $1\times 10^5\text{ Pa}$, but the temperature is decreasing from $500\text{ K}$ to $300\text{ K}$.

Because the temperature is dropping at constant pressure, the volume must also be dropping. The gas is being compressed! $W_{CD}=2R(300-500)=-400R$

The negative sign is crucial here. It physically signifies that work is being done on the gas.

Next, we tackle the vertical lines, which represent processes occurring at a constant temperature. These are isothermal processes.

For process $B\to C$, the temperature is locked at $500\text{ K}$. The pressure drops from $2\times 10^5\text{ Pa}$ to $1\times 10^5\text{ Pa}$. The work done in an isothermal process is: $W=nRT\ln \left(\frac{p_i}{p_f}\right)$

Plugging in our numbers: $W_{BC}=2R(500)\ln \left(\frac{2\times 10^5}{1\times 10^5}\right)=1000R\ln (2)$

Using the approximation $\ln (2)\approx 0.69$, we find: $W_{BC}\approx 1000R\times 0.69=690R$

Similarly, for process $D\to A$, the temperature is constant at $300\text{ K}$, and the pressure increases from $1\times 10^5\text{ Pa}$ to $2\times 10^5\text{ Pa}$. $W_{DA}=2R(300)\ln \left(\frac{1\times 10^5}{2\times 10^5}\right)=600R\ln \left(\frac{1}{2}\right)=-600R\ln (2)$

Again, using our approximation: $W_{DA}\approx -600R\times 0.69=-414R$

We have all the pieces of the puzzle. Now, we just need to sum them up to find the net work done by the gas: $W_{\text{net}}=W_{AB}+W_{BC}+W_{CD}+W_{DA}$

The net work done by the gas is $276R$.

A Word of Caution: If you look at the official answer key or many textbook solutions, you might see the answer listed as $1076R$. How did they get that? They made a fatal physics error! They incorrectly took the work done during the compression process $C\to D$ as positive $400R$ instead of negative $400R$.

Physics is unforgiving when it comes to signs. Compression always results in negative work done by the gas. Trust your concepts, trust the math, and you will see that $276R$ is the undeniable truth!